题目大意:给你一个矩阵,’.’表示空地,’#’表示草地。现在要求你任选两个草地进行点火,使得这个矩阵内的所有草地在最短时间内都着火(如果一个草地着火了,下一秒,和他相邻的草地都会着火)
解题思路:先枚举两个草地,接下来就是bfs判断着火情况了
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
struct Node{
int x, y, time;
Node() {}
Node(int x, int y, int time): x(x), y(y), time(time) {}
};
const int N = 15;
const int INF = 0x3f3f3f3f;
char map[N][N];
bool vis[N][N];
int fire[N * N];
int n, m, cnt, cas = 1;
int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
void init() {
scanf("%d%d", &n, &m);
cnt = 0;
for (int i = 0; i < n; i++) {
scanf("%s", map[i]);
for (int j = 0; j < m; j++)
if (map[i][j] == '#') {
fire[cnt++] = i * m + j;
}
}
}
int bfs(int i, int j) {
memset(vis, 0, sizeof(vis));
queue<Node> Q;
Q.push(Node(fire[i] / m, fire[i] % m, 0));
vis[fire[i] / m][fire[i] % m] = true;
if (i != j) {
Q.push(Node(fire[j] / m, fire[j] % m, 0));
vis[fire[j] / m][fire[j] % m] = true;
}
int t = 0;
while (!Q.empty()) {
Node cur = Q.front(); Q.pop();
for (int i = 0; i < 4; i++) {
int xx = cur.x + dir[i][0];
int yy = cur.y + dir[i][1];
if (xx < 0 || xx >= n || yy < 0 || yy >= m || map[xx][yy] == '.' || vis[xx][yy]) continue;
Q.push(Node(xx, yy, cur.time + 1));
t = cur.time + 1;
vis[xx][yy] = true;
}
}
return t;
}
void solve() {
int ans = INF;
for (int i = 0; i < cnt; i++)
for (int j = i; j < cnt; j++) {
int t = bfs(i, j);
bool flag = false;
for(int i = 0; i < n; i++)
for (int j = 0; j < m; j++) {
if (!(map[i][j] == '.' || vis[i][j])) {
flag = true;
break;
}
}
if (flag == false) ans = min(ans, t);
}
if (ans == INF) ans = -1;
printf("Case %d: %d\n", cas++, ans);
}
int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
solve();
}
return 0;
}